Question

The speech recognition software used by Mr Smiths company can transcribe 130 words per minute. He will purchase a new product only if it can be shown to increase the speed. 41 tests were run on the new software resulting in a mean of 135 and a standard deviation of 4.5 words per minute. With a=.01 should he purchase the new software?

Answer 1

One-Sample t-test

The sample mean is Xˉ=135, the sample standard deviation is s=4.5, and the sample size is n=41. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =130 Ha: μ >130 This corresponds to a Right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2a) Critical Value Based on the information provided, the significance level is α=0.01, and the degree of freedom is n-1=41-1=40. Therefore the critical value for this Right-tailed test is tc​=2.4233. This can be found by either using excel or the t distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is t>2.4233 (3)Test Statistics The t-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that t=7.1146 > tc​=2.4233, it is then concluded that the null hypothesis is rejected. (b) Using p-value method Using the P-value approach: The p-value is p=0, and since p=0≤0.01, it is concluded that the null hypothesis is rejected. (6) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 130, at the 0.01 significance level. He should purchase the new software.